PAT (Advanced Level) Practice 1031 Hello World for U （20 分）-蒲公英云

PAT (Advanced Level) Practice 1031 Hello World for U （20 分）

1031 Hello World for U （20 分）

Given any string of N (≥5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as:

h  d
e  l
l  r
lowo

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible — that is, it must be satisfied that n1=n3=max { k | k≤n2 for all 3≤n2≤N } with n1+n2+n3−2=N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

Sample Output:

h   !
e   d
l   l
lowor

AC代码：

/*1031 Hello World for U
给你一字符串，将它以'U型'输出 
其中n1=n3分别表示U型的左右两侧的字符数，n2等于底部的字符数, N表示字符串里字符的总数 
其中满足关系n1=n3=max{k|k<=n2, 3<=n2<=N}, n1+n2+n3-2=N 
*/
#include <iostream>
#include <string>
using namespace std;
int main(){
    string s;
    cin >> s;
    int N = s.length();
    int n1 = (N + 2) / 3;
    int n2 =  N + 2 - 2*n1;
    //将U型横着输出 
    for(int i = 0; i < n1-1; i++) {
        cout << s[i];
        //输出U中间空的字符串 
        for(int j = 0; j < n2-2; j++){
            cout << " ";
        }
        cout << s[N-i-1] << endl;
    }
    //输出最底部的那一行字符串 
    for(int k = 0; k < n2; k++) {
        cout << s[n1-1+k];
    }
    return 0;
}